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Can we calculate the following integral without the need of $\zeta(2)$, I actually believe that this can be a method to find the accurate value of $\sum_{n\geq 1}n^{-2}$.

$$\int_0^1 -\frac{\log (1-x)}{x} \ \mathrm{d}x$$

Any help is appreciated , just act as you do not know that the sum of reciprocal of squares equal $ \frac{\pi^2}{6}$.

I hope this topic is not duplicated.

Gerry Myerson
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Tulip
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  • Do I understand correctly that you specifically do not want the solution that expresses the function under the integral as a power series, then integrates the summands and finally uses the formula for $\sum_n n^{-2}$ ? – Jakub Konieczny Mar 19 '13 at 20:17
  • If you don't want Riemann's Zeta function, how about Polylogarithm? :) – Kaster Mar 19 '13 at 20:45
  • @Feanor : Yes, I already know that solution, are there any other solutions ? – Tulip Mar 19 '13 at 21:04
  • @Kaster : I know that the indefinite integral is $\text{Li}2(x)$, but the only thing a could get from this is that the integral equals the sum of reciprocal of squares (i.e $\displaystyle \sum{n=0}^{\infty}n^{-2}$) which I can use it in this case (I said just act like you don't know this sum), do you have any ideas ? – Tulip Mar 19 '13 at 21:12

1 Answers1

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$$\int_0^1-\frac{\ln (1-x)}{x}\,dx=\int_0^1\int_0^1\frac{1}{1-xy}\,dy\,dx$$

Now got to page 10 of this article.

L. F.
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